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natka813 [3]
3 years ago
14

Jack researched J.J. Thomson’s experiments. Read Jack’s summary below and find the error.

Chemistry
1 answer:
WINSTONCH [101]3 years ago
8 0

Answer:

D. Error is in sentence 1. The beam was a cathode ray.

Explanation:

Legit just looked up his experiments and it said right off the bat he studied cathode rays not gamma.

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Consider these phylogenetic trees. The first tree is based on physical characteristics. The second tree is based on structure, g
kati45 [8]

Im not 100% sure, but I think the answer is C. If not, Im sorry for bothering you.

7 0
3 years ago
Read 2 more answers
What is the empirical formula of a compound that is 7.74% H and 92.26% C? What is the molecular formula if the molar mass is 78.
Minchanka [31]

Answer:

For all these questions, we want to find the empirical and molecular formulae of various compounds given their percent composition and molar mass. The technique used to answer one of the questions can accordingly be applied to all of them.

Approaching the first question, we treat the percentages of each element as the mass of that element in a 100 g compound (as the percentages add up to 100%). So, our 100 g compound comprises 7.74 g H and 92.26 g C.

Next, we convert these mass quantities into moles. Divide the mass of each element by its molar mass:

7.74 g H/1.00794 g/mol = 7.679 mol H

92.26 g C/12.0107 g/mol = 7.681 mol C.

Then, we look for the molar quantity that's the smallest ("smaller," in this case, since there are only two), and we divide all the molar quantities by the smallest one. Here, it's a very close call, but the number of moles of H is slightly smaller than that of C. So, we divide each molar quantity by the number of moles of H:

7.679 mol H/7.679 mol H = 1

7.681 mol C/7.679 mol H ≈ 1 C/H (the value is actually slightly larger than 1, but we can treat it as 1 for our purposes).

The quotients we calculated represent the subscripts of our compound's empirical formula, which should provide the most simplified whole number ratio of the elements. So the empirical formula of our compound is C₁H₁, or just CH.

Here, it just so happens that we obtained whole number quotients. If we end up with a quotient that isn't a whole number (e.g., 1.5), we would multiply all the quotients by a common number that <em>would </em>give us the most simplified whole number ratio (so, if we had gotten 1 and 1.5, we'd multiply both by 2, and the empirical formula would have subscripts 2 and 3).

To find the molecular formula (the actual formula of our compound), we use the molar mass of the compound, 78.1134 g/mol. The molar mass of our "empirical compound," CH, is 13.0186 g/mol. Since our empirical formula represents the most simplified molar ratio of the elements, the molar masses of our "empirical compound" and the actual compound should be multiples of one another. We divide 78.1134 g/mol by 13.0176 g/mol and obtain 6. The subscripts in our molecular formula are equal to the subscripts in our empirical formula multiplied by 6.

Thus, our molecular formula is C₆H₆.

---

As mentioned before, all the questions here can be answered following the procedure used to answer the first question above. In any case, I've provided the empirical and molecular formulae for the remaining questions below for your reference.

2. Empirical formula: C₁₃H₁₂O; molecular formula: C₁₃H₁₂O

3. Empirical formula: CH; molecular formula: C₈H₈

4. Empirical formula: C₂HCl; molecular formula: C₆H₃Cl₃

5. Empirical formula: Cl₄K₂Pt; molecular formula: Cl₄K₂Pt

6. Empirical formula: C₂H₄Cl; molecular formula: C₄H₈Cl₂

6 0
2 years ago
Explain how materials are suited for different uses based on their physical and chemical properties?
Leya [2.2K]
Ok cool I ujust do you think I should be able to get a hold with him tomorrow morning or t if rry want me too early or something I don’t know how to get it off or if you want me too I just need it for a while and I’ll get back with him
4 0
2 years ago
In a certain city, electricity costs $0.17 per kW·h. What is the annual cost for electricity to power a lamp-post for 5.50 hours
Anon25 [30]

Answer:

(a) = $34.123

(b) = $8.532

(c) Additional cost of fluorescent bulb is justified

Explanation:

Cost of electricity = $0.17 per kW·h

(a) For a 100 watt bulb which is the same as 100/1000 or 0.1 kW, the cost per hour =

0.1 × 0.17 = $0.017/h

and for 5.5 hours = 0.017×5.5 = $0.0935

The annual cost, which is 365 days, we have

Annual cost = $0.0935 × 365 = $34.123

(b) For the energy efficient 25-watt bulb, we have

25/1000 = 0.025kW

Power cost per annum =

0.025kW×$0.17 per kW·h×5.5×365 = $8.532

(c) Total cost of incandescent bulb = $0.89 total cost of using the incandescent bulb is $34.123 + $0.89 = $35.02

Total cost of using the energy efficient fluorescent bulb is about $3.49

Total cost of using the energy efficient bulb = $8.532 + $3.49 = $12.02

Total cost of incandescent bulb = $35.02 while total cost of energy efficient bulb is = $12.02

$12.02 <$35.02

Additional cost of fluorescent bulb is justified

8 0
3 years ago
Calculate the number of calories needed to increase the temperature of 50.0 g of copper metal from 21.0 degrees C to 75.0 degree
KonstantinChe [14]
<h3>Answer:</h3>

1031.4 Calories.

<h3>Explanation:</h3>

We are given;

Mass of the copper metal = 50.0 g

Initial temperature = 21.0 °C

Final temperature, = 75°C

Change in temperature = 54°C

Specific heat capacity of copper = 0.382 Cal/g°C

We are required to calculate the amount of heat in calories required to raise the temperature of the copper metal;

Quantity of heat is given by the formula,

Q = Mass × specific heat capacity × change in temperature

   = 50.0 g × 0.382 Cal/g°C × 54 °C

   = 1031.4 Calories

Thus, the amount of heat energy required is 1031.4 Calories.

4 0
3 years ago
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